Question

Two large vertical and parallel metal plates having a separation of 1cm are connected to a DC voltage source of potential difference X. A proton is released at rest midway between the two plates. It remains at rest in the air then X is:
\[\begin{align}
  & A)\text{ 1}\times \text{1}{{\text{0}}^{-5}}V \\
 & B)\text{ 1}\times \text{1}{{\text{0}}^{-7}}V \\
 & C)\text{ 1}\times \text{1}{{\text{0}}^{-9}}V \\
 & D)\text{ 1}\times \text{1}{{\text{0}}^{-10}}V \\
\end{align}\]

Answer 1

Hint: Two large vertical parallel metal plates separated by a distance is the elaborated definition for a capacitor. So, we have to use the capacitance formulae and concepts to solve the Voltage between the plates. The release of a proton indicates that the air between the gap has ionized.

Complete answer:
Let us consider a capacitor with separation of 1cm and dielectric medium as air.
            

We know that the release of proton was a result of the energy produced by the DC source which created an electric field enough for the proton to be ionised.
We can calculate the equivalent force required to ionise a proton of mass ‘m’. We know that the force developed by an electric field if
\[F=qE\]
The force required to ionise a particle of mass ‘m’ is equivalent to its force due to mass, i.e.,
\[{{F}^{'}}=mg\]
Now, at the time of ionisation, these two forces must be equal, which will give the Electric field strength required for ionisation. i.e.,
\[qE=mg\]
Now we know that the potential difference X is related to the electric field as:
\[\begin{align}
  & \Rightarrow E=\dfrac{V}{d} \\
 & \Rightarrow V=Ed \\
\end{align}\]
Given that,
\[\begin{align}
  & d=1cm={{10}^{-2}}m \\
 & m={{m}_{p}}=1.6\times {{10}^{-27}}kg \\
 & q=1.6\times {{10}^{-19}}C \\
 & g=10m{{s}^{-2}} \\
\end{align}\]
Now equating the forces,
\[\begin{align}
  & \Rightarrow F={{F}^{'}} \\
 & \Rightarrow qE=mg \\
 & \Rightarrow q\dfrac{V}{d}=mg \\
 & \Rightarrow V=\dfrac{mgd}{q} \\
 & \Rightarrow V=\dfrac{1.6\times {{10}^{-27}}\times 10\times {{10}^{-2}}}{1.6\times {{10}^{-19}}} \\
 & \Rightarrow V=1\times {{10}^{-9}}V \\
\end{align}\]
i.e., the required voltage X is \[1\times {{10}^{-9}}V\].

The correct answer is option C.

Note:
The mass of the proton is approximated to two digits for the ease of solving. The need to consider the situation as a capacitor is based on the reader’s way of understanding. It can be considered a direct problem as we have done above or can be more related to a capacitor.